How to show that a function has slow or rapid decay I've got a function whose general form is
$f(x) = \frac{1}{x^\alpha}$
where $x > 0$ and $\alpha > 0$. I would like to show that if $0< \alpha < 1$ $f(x)$ has slow decay and if $\alpha > 1$ the $f(x)$ has rapid decay. (I've already verified these properties of $f(x)$ using a graphing application.)
 A: If you start at 1 (your sum is not defined at x=0), you can bound the sums as integrals: $ \int_{x=2}^{\infty}x^{-\alpha} dx \lt \sum_{x = 1}^\infty x^{-\alpha} \lt \int_{x=1}^{\infty}x^{-\alpha} dx$.  One will solve it for $\alpha \le 1$ and one for $\alpha \gt 1$
A: Another approach is to use
Cauchy's condensation test
which states that a series $\sum a_n$, where the terms $a_n$
are positive and decreasing, is convergent if and only if
$\sum 2^m a_{2^m}$ is convergent.
In this example the line $\mathbf{Re}(s)=1$ is the
edge of the region of ansolute convergence of the usual
formula $\sum_{n=1}^\infty 1/n^s$ for the Riemann zeta function.
A: I'm probably talking nonsense but 
$\int_1^\infty \frac{1}{x^\alpha} = \left. \frac{x^{1 - \alpha}}{1 - \alpha} \right|_1^\infty$
The integral diverges if $1 - \alpha > 0$ i.e. if $\alpha < 1$, but converges if $1 - \alpha < 0$ i.e. $\alpha > 1$.
My difficulty now is in showing that the integral is a bound on the sum, which from what I've studied can only be done if the terms of the sum are taken as follows
 
instead of 
 
which is more natural and intuitive.
